我有一个带有虚拟属性的基类,并且派生类型覆盖了该虚拟属性。该类型可以序列化为XML。我试图做的是当对象属于派生类型时,不保留项目列表属性。为此,派生类使用[XmlIgnore]
属性装饰了重写的属性。基类中的虚拟属性不适用于XmlIgnore
属性。由于某种原因,即使对象是派生类型(DynamicCart
),项目列表也会被序列化。
当我将XmlIgnore
属性应用于基类中的虚拟属性时,列表不会序列化到文件。
public class ShoppingCart
{
public virtual List<items> Items{get; set;}
//and other properties
public void SerializeToXML (string filePath)
{
var xmlSerializer = new XmlSerializer(this.GetType());
textWriter = new System.IO.StreamWriter(filePath);
xmlSerializer.Serialize(textWriter, this);
textWriter.Flush();
textWriter.Close();
}
}
//A cart that is populated by algo based on parameters supplied by user. I have no need to
//persist the actual items across sessions.
class DynamicCart: ShoppingCart
{
[XmlIgnore]
public override List<items>{get;set;}
//and other properties
}
class Shop
{
ShoppingCart cart = new DynamicCart();
PopulateCart(cart);
cart.serializeToXML(<PATH TO FILE>);
}
最佳答案
您可以通过在基类中添加虚拟ShouldSerialize***
方法来实现。例如:
[XmlInclude(typeof(Sub))]
public class Base
{
public virtual string Prop { get; set; }
public virtual bool ShouldSerializeProp() { return true; }
}
public class Sub : Base
{
public override string Prop { get; set; }
public override bool ShouldSerializeProp() { return false; }
}
internal class Program
{
private static void Main()
{
var o = new Sub { Prop = "Value" };
var baseSer = new XmlSerializer(typeof (Base));
var subSer = new XmlSerializer(typeof (Sub));
Console.Out.WriteLine("BASE:");
baseSer.Serialize(Console.Out, o);
Console.Out.WriteLine();
Console.Out.WriteLine("SUB:");
subSer.Serialize(Console.Out, o);
Console.Out.WriteLine();
Console.ReadLine();
}
}
这产生(稍微整理):
BASE:
<Base xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:type="Sub">
<Prop>Value</Prop>
</Base>
SUB:
<Sub />
该方法必须在
ShouldInclude...
之后包含要考虑的属性的确切名称。